The polar of $(- 3, 5)$ w.r.t $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ cuts the circle, its mid point is

(\frac{- 17}{25}, \frac{56}{25})

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(\frac{- 17}{25}, \frac{- 56}{25})

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(\frac{17}{25}, \frac{56}{25})

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(\frac{17}{25}, \frac{- 56}{25})

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Solution

The correct option is C $(\frac{17}{25}, \frac{56}{25})$
Polar of $(- 3, 5)$ wrt $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ is chord of contact from $(- 3, 5)$ to a given circle.
$∴ x (- 3) + 5 y - (x - 3) - 2 (y + 5) + 3 = 0$
$- 4 x + 3 y - 4 = 0$
$4 x - 3 y + 4 = 0$ -----(1)
mid-point of polar of $(- 3, 5)$ is foot perpendicular to chord $4 x - 3 y + 4 = 0$ from centre of circle
$\frac{x - 1}{4} = \frac{y - 2}{- 3} = - \frac{(2)}{4^{2} + 3^{2}}$
$x = 1 - \frac{8}{25} = \frac{17}{25}$
and $y = 2 + \frac{6}{25} = \frac{65}{25}$
$\Rightarrow (\frac{17}{25}, \frac{56}{25})$ is a mid-point of polar of $(- 3, 5)$

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Question 17

$(- 25) \times [6 + 4]$ is not same as

(a) $(- 25) \times 10$
(b) $(- 25) \times 6 + (- 25) \times (4)$
(c) $(- 25) \times (6) \times (4)$
(d) -250

3 x + 4 y = 25

x^{2} + y^{2} = 25

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